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Give the n factors for the following changes: -
(i) $ N{a_2}C{O_3} \to NaHC{O_3} $
(ii) $ A{l_2}{\left( {S{O_4}} \right)_3} \to Al{\left( {OH} \right)_3} $
(iii) $ N{a_2}{S_2}{O_3} \to {S_4}O_6^{2 - } $
(A) $ 1,3,2 $
(B) $ 1,3,1 $
(C) $ 1,6,1 $
(D) $ 2,6,1 $

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Last updated date: 24th Jul 2024
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Answer
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Hint :The $ N{a_2}C{O_3} $ is sodium carbonate, which is being converted in sodium bicarbonate $ \left( {NaHC{O_3}} \right). $ $ A{l_2}{\left( {S{O_4}} \right)_3} $ is aluminum sulphate, and is being converted to aluminum hydroxide $ \left( {Al{{\left( {OH} \right)}_3}} \right). $ $ N{a_2}{S_2}{O_3} $ is sodium thiosulphate is converted to tetrathionate anion $ \left( {{S_4}O_6^{2 - }} \right). $

Complete Step By Step Answer:
We have:
(i) $ N{a_2}C{O_3} \to NaHC{O_3} $
For $ N{a_2}C{O_3} \to NaHC{O_3} $ the total change in the positive charge of the cations in the compound is the value of the n-factor. In $ N{a_2}C{O_3}, $ carbonate anion $ \left( {CO_3^{2 - }} \right) $ has $ - 2 $ charge. This means that the total positive charge contributed by the cations $ \left( {2N{a^ + }} \right) $ is $ + 2. $ In $ NaHC{O_3}, $ bicarbonate anion has $ - 1 $ charge, and hence total positive charge contributed by cation $ \left( {N{a^ + }} \right) $ is $ + 1. $
Therefore, n-factor value $ = 2 - 1 = 1 $

(ii) $ A{l_2}{\left( {S{O_4}} \right)_3} \to Al{\left( {OH} \right)_3} $
For $ A{l_2}{\left( {S{O_4}} \right)_3} \to Al{\left( {OH} \right)_3} $ the total change in the positive charge of the cations in the compound is the value of the n-factor. In $ A{l_2}{\left( {S{O_4}} \right)_3}, $ sulphate anion $ \left( {SO_4^{2 - }} \right) $ has $ - 2 $ charge. The total negative charge will therefore be $ - 2 \times 3 = 6 $ as there are three sulphate ions. This means that the total positive charge contributed by the cations $ \left( {2A{l^{3 + }}} \right) $ is $ + 6. $ In $ Al{\left( {OH} \right)_3}, $ hydroxide ion has $ - 1 $ charge, and hence three hydroxide ions contribute $ - 3 $ charge. The total positive charge contributed by cation $ \left( {A{l^{3 + }}} \right) $ is $ + 3. $
Therefore, n-factor value $ = 6 - 3 = 3 $
(iii) $ N{a_2}{S_2}{O_3} \to {S_4}O_6^{2 - } $
For $ N{a_2}{S_2}{O_3} \to {S_4}O_6^{2 - } $ the n-factor is the product of change in oxidation state and number of atoms undergoing change. Since the change undergone is as follows: $ {S_2}O_3^{2 - } \to {S_4}O_6^{2 - } $ we need to calculate the oxidation state of sulphur atom in each of these. The oxidation states of sulphur in $ {S_2}O_3^{2 - } $ and $ {S_4}O_6^{2 - } $ are $ 2 $ and $ 2.5 $ respectively and hence the change in oxidation state is $ 2.5 - 2 = 0.5. $ The number of sulphur atoms undergoing change are $ 4 - 2 = 2. $
Therefore, n-factor value $ = 0.5 \times 2 = 1 $
Hence, the required answer is (B) $ 1,3,1. $

Note :
The n-factor can be calculated for different compounds according to their characteristics: whether they are an acid, base, salt or if the reaction undergone by the compound is a redox reaction. For acids, the n-factor is represented in the form of basicity, while for bases, the n-factor is represented in the form of acidity. For salts, the number of positive or negative ions are counted. For redox reactions, the change in oxidation number or change in their reduction number in both sides of the reaction is considered.